Integrand size = 20, antiderivative size = 141 \[ \int e^{3 \text {arctanh}(a x)} (c-a c x)^{9/2} \, dx=\frac {256 c^7 \left (1-a^2 x^2\right )^{5/2}}{1155 a (c-a c x)^{5/2}}+\frac {64 c^6 \left (1-a^2 x^2\right )^{5/2}}{231 a (c-a c x)^{3/2}}+\frac {8 c^5 \left (1-a^2 x^2\right )^{5/2}}{33 a \sqrt {c-a c x}}+\frac {2 c^4 \sqrt {c-a c x} \left (1-a^2 x^2\right )^{5/2}}{11 a} \] Output:
256/1155*c^7*(-a^2*x^2+1)^(5/2)/a/(-a*c*x+c)^(5/2)+64/231*c^6*(-a^2*x^2+1) ^(5/2)/a/(-a*c*x+c)^(3/2)+8/33*c^5*(-a^2*x^2+1)^(5/2)/a/(-a*c*x+c)^(1/2)+2 /11*c^4*(-a*c*x+c)^(1/2)*(-a^2*x^2+1)^(5/2)/a
Time = 0.06 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.44 \[ \int e^{3 \text {arctanh}(a x)} (c-a c x)^{9/2} \, dx=-\frac {2 c^4 (1+a x)^{5/2} \sqrt {c-a c x} \left (-533+755 a x-455 a^2 x^2+105 a^3 x^3\right )}{1155 a \sqrt {1-a x}} \] Input:
Integrate[E^(3*ArcTanh[a*x])*(c - a*c*x)^(9/2),x]
Output:
(-2*c^4*(1 + a*x)^(5/2)*Sqrt[c - a*c*x]*(-533 + 755*a*x - 455*a^2*x^2 + 10 5*a^3*x^3))/(1155*a*Sqrt[1 - a*x])
Time = 0.53 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6677, 459, 459, 459, 458}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int e^{3 \text {arctanh}(a x)} (c-a c x)^{9/2} \, dx\) |
\(\Big \downarrow \) 6677 |
\(\displaystyle c^3 \int (c-a c x)^{3/2} \left (1-a^2 x^2\right )^{3/2}dx\) |
\(\Big \downarrow \) 459 |
\(\displaystyle c^3 \left (\frac {12}{11} c \int \sqrt {c-a c x} \left (1-a^2 x^2\right )^{3/2}dx+\frac {2 c \left (1-a^2 x^2\right )^{5/2} \sqrt {c-a c x}}{11 a}\right )\) |
\(\Big \downarrow \) 459 |
\(\displaystyle c^3 \left (\frac {12}{11} c \left (\frac {8}{9} c \int \frac {\left (1-a^2 x^2\right )^{3/2}}{\sqrt {c-a c x}}dx+\frac {2 c \left (1-a^2 x^2\right )^{5/2}}{9 a \sqrt {c-a c x}}\right )+\frac {2 c \left (1-a^2 x^2\right )^{5/2} \sqrt {c-a c x}}{11 a}\right )\) |
\(\Big \downarrow \) 459 |
\(\displaystyle c^3 \left (\frac {12}{11} c \left (\frac {8}{9} c \left (\frac {4}{7} c \int \frac {\left (1-a^2 x^2\right )^{3/2}}{(c-a c x)^{3/2}}dx+\frac {2 c \left (1-a^2 x^2\right )^{5/2}}{7 a (c-a c x)^{3/2}}\right )+\frac {2 c \left (1-a^2 x^2\right )^{5/2}}{9 a \sqrt {c-a c x}}\right )+\frac {2 c \left (1-a^2 x^2\right )^{5/2} \sqrt {c-a c x}}{11 a}\right )\) |
\(\Big \downarrow \) 458 |
\(\displaystyle c^3 \left (\frac {12}{11} c \left (\frac {8}{9} c \left (\frac {8 c^2 \left (1-a^2 x^2\right )^{5/2}}{35 a (c-a c x)^{5/2}}+\frac {2 c \left (1-a^2 x^2\right )^{5/2}}{7 a (c-a c x)^{3/2}}\right )+\frac {2 c \left (1-a^2 x^2\right )^{5/2}}{9 a \sqrt {c-a c x}}\right )+\frac {2 c \left (1-a^2 x^2\right )^{5/2} \sqrt {c-a c x}}{11 a}\right )\) |
Input:
Int[E^(3*ArcTanh[a*x])*(c - a*c*x)^(9/2),x]
Output:
c^3*((2*c*Sqrt[c - a*c*x]*(1 - a^2*x^2)^(5/2))/(11*a) + (12*c*((2*c*(1 - a ^2*x^2)^(5/2))/(9*a*Sqrt[c - a*c*x]) + (8*c*((8*c^2*(1 - a^2*x^2)^(5/2))/( 35*a*(c - a*c*x)^(5/2)) + (2*c*(1 - a^2*x^2)^(5/2))/(7*a*(c - a*c*x)^(3/2) )))/9))/11)
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(p + 1))), x] /; FreeQ[{a, b, c , d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && EqQ[n + p, 0]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[2*c* (Simplify[n + p]/(n + 2*p + 1)) Int[(c + d*x)^(n - 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && IGtQ[Simplif y[n + p], 0]
Int[E^(ArcTanh[(a_.)*(x_)]*(n_.))*((c_) + (d_.)*(x_))^(p_.), x_Symbol] :> S imp[c^n Int[(c + d*x)^(p - n)*(1 - a^2*x^2)^(n/2), x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[a*c + d, 0] && IntegerQ[(n - 1)/2] && IntegerQ[2*p]
Time = 0.10 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.45
method | result | size |
gosper | \(\frac {2 \left (a x +1\right )^{4} \left (105 a^{3} x^{3}-455 a^{2} x^{2}+755 a x -533\right ) \left (-a c x +c \right )^{\frac {9}{2}}}{1155 a \left (a x -1\right )^{3} \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}\) | \(63\) |
orering | \(\frac {2 \left (a x +1\right )^{4} \left (105 a^{3} x^{3}-455 a^{2} x^{2}+755 a x -533\right ) \left (-a c x +c \right )^{\frac {9}{2}}}{1155 a \left (a x -1\right )^{3} \left (-a^{2} x^{2}+1\right )^{\frac {3}{2}}}\) | \(63\) |
default | \(\frac {2 \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, c^{4} \left (a x +1\right )^{2} \left (105 a^{3} x^{3}-455 a^{2} x^{2}+755 a x -533\right )}{1155 \left (a x -1\right ) a}\) | \(67\) |
risch | \(\frac {2 \sqrt {-\frac {\left (-a^{2} x^{2}+1\right ) c}{a x -1}}\, \left (a x -1\right ) c^{5} \left (105 a^{5} x^{5}-245 a^{4} x^{4}-50 a^{3} x^{3}+522 a^{2} x^{2}-311 a x -533\right ) \left (a x +1\right )}{1155 \sqrt {-a^{2} x^{2}+1}\, \sqrt {-c \left (a x -1\right )}\, a \sqrt {c \left (a x +1\right )}}\) | \(110\) |
Input:
int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a*c*x+c)^(9/2),x,method=_RETURNVERBOSE)
Output:
2/1155*(a*x+1)^4*(105*a^3*x^3-455*a^2*x^2+755*a*x-533)*(-a*c*x+c)^(9/2)/a/ (a*x-1)^3/(-a^2*x^2+1)^(3/2)
Time = 0.07 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.65 \[ \int e^{3 \text {arctanh}(a x)} (c-a c x)^{9/2} \, dx=\frac {2 \, {\left (105 \, a^{5} c^{4} x^{5} - 245 \, a^{4} c^{4} x^{4} - 50 \, a^{3} c^{4} x^{3} + 522 \, a^{2} c^{4} x^{2} - 311 \, a c^{4} x - 533 \, c^{4}\right )} \sqrt {-a^{2} x^{2} + 1} \sqrt {-a c x + c}}{1155 \, {\left (a^{2} x - a\right )}} \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a*c*x+c)^(9/2),x, algorithm="fric as")
Output:
2/1155*(105*a^5*c^4*x^5 - 245*a^4*c^4*x^4 - 50*a^3*c^4*x^3 + 522*a^2*c^4*x ^2 - 311*a*c^4*x - 533*c^4)*sqrt(-a^2*x^2 + 1)*sqrt(-a*c*x + c)/(a^2*x - a )
\[ \int e^{3 \text {arctanh}(a x)} (c-a c x)^{9/2} \, dx=\int \frac {\left (- c \left (a x - 1\right )\right )^{\frac {9}{2}} \left (a x + 1\right )^{3}}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((a*x+1)**3/(-a**2*x**2+1)**(3/2)*(-a*c*x+c)**(9/2),x)
Output:
Integral((-c*(a*x - 1))**(9/2)*(a*x + 1)**3/(-(a*x - 1)*(a*x + 1))**(3/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 254 vs. \(2 (117) = 234\).
Time = 0.08 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.80 \[ \int e^{3 \text {arctanh}(a x)} (c-a c x)^{9/2} \, dx=-\frac {2 \, {\left (35 \, a^{6} c^{\frac {9}{2}} x^{6} - 175 \, a^{5} c^{\frac {9}{2}} x^{5} + 415 \, a^{4} c^{\frac {9}{2}} x^{4} - 741 \, a^{3} c^{\frac {9}{2}} x^{3} + 1482 \, a^{2} c^{\frac {9}{2}} x^{2} - 5928 \, a c^{\frac {9}{2}} x - 11856 \, c^{\frac {9}{2}}\right )}}{385 \, \sqrt {a x + 1} a} - \frac {2 \, {\left (7 \, a^{5} c^{\frac {9}{2}} x^{5} - 37 \, a^{4} c^{\frac {9}{2}} x^{4} + 97 \, a^{3} c^{\frac {9}{2}} x^{3} - 215 \, a^{2} c^{\frac {9}{2}} x^{2} + 860 \, a c^{\frac {9}{2}} x + 1720 \, c^{\frac {9}{2}}\right )}}{21 \, \sqrt {a x + 1} a} - \frac {6 \, {\left (5 \, a^{4} c^{\frac {9}{2}} x^{4} - 29 \, a^{3} c^{\frac {9}{2}} x^{3} + 93 \, a^{2} c^{\frac {9}{2}} x^{2} - 407 \, a c^{\frac {9}{2}} x - 814 \, c^{\frac {9}{2}}\right )}}{35 \, \sqrt {a x + 1} a} - \frac {2 \, {\left (a^{3} c^{\frac {9}{2}} x^{3} - 7 \, a^{2} c^{\frac {9}{2}} x^{2} + 43 \, a c^{\frac {9}{2}} x + 91 \, c^{\frac {9}{2}}\right )}}{5 \, \sqrt {a x + 1} a} \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a*c*x+c)^(9/2),x, algorithm="maxi ma")
Output:
-2/385*(35*a^6*c^(9/2)*x^6 - 175*a^5*c^(9/2)*x^5 + 415*a^4*c^(9/2)*x^4 - 7 41*a^3*c^(9/2)*x^3 + 1482*a^2*c^(9/2)*x^2 - 5928*a*c^(9/2)*x - 11856*c^(9/ 2))/(sqrt(a*x + 1)*a) - 2/21*(7*a^5*c^(9/2)*x^5 - 37*a^4*c^(9/2)*x^4 + 97* a^3*c^(9/2)*x^3 - 215*a^2*c^(9/2)*x^2 + 860*a*c^(9/2)*x + 1720*c^(9/2))/(s qrt(a*x + 1)*a) - 6/35*(5*a^4*c^(9/2)*x^4 - 29*a^3*c^(9/2)*x^3 + 93*a^2*c^ (9/2)*x^2 - 407*a*c^(9/2)*x - 814*c^(9/2))/(sqrt(a*x + 1)*a) - 2/5*(a^3*c^ (9/2)*x^3 - 7*a^2*c^(9/2)*x^2 + 43*a*c^(9/2)*x + 91*c^(9/2))/(sqrt(a*x + 1 )*a)
Time = 0.18 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.52 \[ \int e^{3 \text {arctanh}(a x)} (c-a c x)^{9/2} \, dx=-\frac {2 \, {\left (512 \, \sqrt {2} c^{\frac {7}{2}} + \frac {105 \, {\left (a c x + c\right )}^{\frac {11}{2}} - 770 \, {\left (a c x + c\right )}^{\frac {9}{2}} c + 1980 \, {\left (a c x + c\right )}^{\frac {7}{2}} c^{2} - 1848 \, {\left (a c x + c\right )}^{\frac {5}{2}} c^{3}}{c^{2}}\right )} c^{2}}{1155 \, a {\left | c \right |}} \] Input:
integrate((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a*c*x+c)^(9/2),x, algorithm="giac ")
Output:
-2/1155*(512*sqrt(2)*c^(7/2) + (105*(a*c*x + c)^(11/2) - 770*(a*c*x + c)^( 9/2)*c + 1980*(a*c*x + c)^(7/2)*c^2 - 1848*(a*c*x + c)^(5/2)*c^3)/c^2)*c^2 /(a*abs(c))
Time = 22.71 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.64 \[ \int e^{3 \text {arctanh}(a x)} (c-a c x)^{9/2} \, dx=\frac {\sqrt {c-a\,c\,x}\,\left (\frac {1688\,c^4\,x}{1155}+\frac {1066\,c^4}{1155\,a}-\frac {422\,a\,c^4\,x^2}{1155}-\frac {944\,a^2\,c^4\,x^3}{1155}+\frac {118\,a^3\,c^4\,x^4}{231}+\frac {8\,a^4\,c^4\,x^5}{33}-\frac {2\,a^5\,c^4\,x^6}{11}\right )}{\sqrt {1-a^2\,x^2}} \] Input:
int(((c - a*c*x)^(9/2)*(a*x + 1)^3)/(1 - a^2*x^2)^(3/2),x)
Output:
((c - a*c*x)^(1/2)*((1688*c^4*x)/1155 + (1066*c^4)/(1155*a) - (422*a*c^4*x ^2)/1155 - (944*a^2*c^4*x^3)/1155 + (118*a^3*c^4*x^4)/231 + (8*a^4*c^4*x^5 )/33 - (2*a^5*c^4*x^6)/11))/(1 - a^2*x^2)^(1/2)
Time = 0.16 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.38 \[ \int e^{3 \text {arctanh}(a x)} (c-a c x)^{9/2} \, dx=\frac {2 \sqrt {c}\, \sqrt {a x +1}\, c^{4} \left (-105 a^{5} x^{5}+245 a^{4} x^{4}+50 a^{3} x^{3}-522 a^{2} x^{2}+311 a x +533\right )}{1155 a} \] Input:
int((a*x+1)^3/(-a^2*x^2+1)^(3/2)*(-a*c*x+c)^(9/2),x)
Output:
(2*sqrt(c)*sqrt(a*x + 1)*c**4*( - 105*a**5*x**5 + 245*a**4*x**4 + 50*a**3* x**3 - 522*a**2*x**2 + 311*a*x + 533))/(1155*a)